Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{k - 7}{k - 1} \times \dfrac{k^2 - k}{k^2 + 3k - 70} $
Solution: First factor the quadratic. $a = \dfrac{k - 7}{k - 1} \times \dfrac{k^2 - k}{(k - 7)(k + 10)} $ Then factor out any other terms. $a = \dfrac{k - 7}{k - 1} \times \dfrac{k(k - 1)}{(k - 7)(k + 10)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (k - 7) \times k(k - 1) } { (k - 1) \times (k - 7)(k + 10) } $ $a = \dfrac{ k(k - 7)(k - 1)}{ (k - 1)(k - 7)(k + 10)} $ Notice that $(k - 1)$ and $(k - 7)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ k\cancel{(k - 7)}(k - 1)}{ (k - 1)\cancel{(k - 7)}(k + 10)} $ We are dividing by $k - 7$ , so $k - 7 \neq 0$ Therefore, $k \neq 7$ $a = \dfrac{ k\cancel{(k - 7)}\cancel{(k - 1)}}{ \cancel{(k - 1)}\cancel{(k - 7)}(k + 10)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $a = \dfrac{k}{k + 10} ; \space k \neq 7 ; \space k \neq 1 $